(1) Reduction to slope - intercept form:

(2) Reduction to intercept form:

(3) Reduction to normal form:

$Ax +By + c =0$

$\Rightarrow By =-Ax-c$

$\Rightarrow y=-\frac{A}{B}x-\frac{c}{B}$

$m=-\frac{A}{B}$ & $c=-\frac{c}{B}$

$y = mx+c$

$A x+B y+C=0$

$\Rightarrow A x+B y=-C \quad \frac{x}{a}+\frac{y}{b}=1$

$\Rightarrow -\frac{A}{C} x-\frac{B}{c} y=1$

$\Rightarrow \frac{x}{-C / A}+\frac{y}{-C / B}=1$

$a=-C / A \quad$ & $b=-C / B$

$A x+B y+C=0$

$x \cos \alpha+y \sin \alpha-p=0$

$\frac{A}{\cos \alpha}=\frac{B}{\sin \alpha}=-\frac{C}{p}=k(\operatorname{say})$

$\Rightarrow A=k \cos \alpha, B=k \sin \alpha, C=-p k$

$\Rightarrow A^2+B^2=k^2\left(\cos ^2 \alpha+\sin ^2 \alpha\right)$

$=k^2$

$\therefore k= \pm \sqrt{A^2+B^2}$

$c=-p k \Rightarrow p=-\frac{c}{k}=\frac{-c}{ \pm \sqrt{A^2+B^2}}$

cone (i) When $c<0$, then

$p=\frac{c}{-\sqrt{A^2+B^2}}$

(ii) When $c>0$, then

$p=\frac{c}{+\sqrt{A^2+B^2}}$

$A x+B y+c=0 \Rightarrow \pm \frac{A}{\sqrt{A+B^2}} x \pm \frac{B}{\sqrt{A+B}{ }^2}$ $=\frac{c}{\sqrt{A+B}}$

Examples on Various forms of equation of straight lines :

1. Find the equation of the straight line through the point $(-1,-2)$ making an angle of $135^0$ with the $x-$ axis.

Solution:

Give $\theta=135^0 \Rightarrow m= \tan 135^0=-1$

line passing through $(-1,-2)$

$\therefore \quad\left(y-y_1\right)=m\left(x-x_1\right)$

$\Rightarrow \quad(y+2)=-1(x+1)$ $\Rightarrow \quad x+y+3=0$

Find the equation of the line passing through $(2,3)$ and making equal intercepts on the coordinate axes.

Solution:

Let intercepts are $a$ & $a$

$\frac{x}{a}+\frac{y}{a}= 1$

$\Rightarrow x + y = a\rightarrow l$

line l passing through $(2,3)$

$2 + 3 =a \Rightarrow a=5$

$\frac{x}{5} + \frac{y}{5} =1 \Rightarrow 2 + y = 5$

Find the equation of the line passing through the points $(1,2)$ and $(0,5).$

Solution:

Give line l passing through $A(1,2)$ & $B(0,5)$.. Equation of line $AB$ slope of line $AB=\frac{5-2}{0-1}= - 3$

Equation of line AB passing though A(1,2) with slope $= -3$ is

$(y-y_1) = m (x-x_1)$

$\Rightarrow (y-2) = -3 (x-1)$

$\Rightarrow 3x+y-5=0$

Determine the equation of the line with $\alpha=135^{\circ}$ and perpendicular distance $p=\sqrt{2}$ from the origin.
Solution:

Given $\alpha=135^{\circ}$ & $p=\sqrt{2}$
Equation of line $l,$ $\ x \cos \alpha+y \sin \alpha=p$

$\Rightarrow x \cos 135^{\circ}+y \sin 135^{\circ}=p$

$\Rightarrow-\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\sqrt{2}$

$\Rightarrow-x+y=2$ $\Rightarrow x-y+2=0$

Find the angle between the lines $3x+y-7=0$ & $x+2y+9=0.$

Solution:

Given $l_1 : 3x+y-7=0$ & $l_2 : x+2y+9=0$

$\therefore m_1=-3$ & $m_2=-\frac{1}{2}$

Let $\theta$ be the angle between $l_1$ & $l_2$

$\therefore \tan \theta =\left|\frac{m_1-m_2}{1+m_1 m_2}\right| =\left|\frac{-3+\frac{1}{2}}{1+3 \times \frac{1}{2}}\right|=\left|\frac{-5 / 2}{5 / 2}\right| = \pm 1$

$\therefore$ when $\tan\theta=1 \ \Rightarrow \theta =\frac{\pi}{4}\qquad\quad$ when $\tan \theta = -1 \ \Rightarrow \theta = \frac{3\pi}{4}$

Reduce the equation $x+\sqrt 3y+4=0$ to perpendicular form and hence find the lenght of the perpendicular from the origin on the straight line.

Solution:

Given equation $x+\sqrt{3} y+4=0$

$\therefore A=1, B=\sqrt{3}, C=4$

$\therefore \sqrt{A^2+B^2}=\sqrt{1+3}=\sqrt{4}= \pm 2 \quad\qquad$ Since $c>0 \Rightarrow \sqrt{A^2 + B^2}=2$

Equation in normal then

$\frac{A}{\sqrt{A^2+ B^2}} x+\frac{B}{\sqrt{A^2+B^2}} y=\frac{C}{\sqrt{A^2+ B^2}}$

$\Rightarrow \quad \frac{1}{2} x+\frac{\sqrt{3}}{2} y=\frac{4}{2}$

$\Rightarrow \quad x \cos \frac{\pi}{3}+y \sin \frac{\pi}{3}=2$

$\therefore \alpha=\frac{\pi}{3}$ & $\quad p=2$

Perpendicular distance of line from origin.

Find the equation of the stright line which passing through the point (-1,3) and is perpendicular to the line $4x+3y+1=0.$

Solution:

Given equation of line $l_1: 4x+3y+1=0$

$\therefore$ slope of $l_1(m_1)=-\frac{4}{3}$

Since $l_1 \text{ perpendiculer } l_2$

$m_1 \times m_2= -1$

$m_2=-\frac{1}{m_1}=\frac{3}{4}$

Equation of line $l_1$ passing through $l(-1,3)$ is

$(y-3) =\frac{3}{4} (x+1)$

$\Rightarrow 4y-12 = 3x + 3 \Rightarrow 3x-4y + 15=0$

Find the equation of a line which passes through the point (3,1) and bisect the portion of the line $3x+4y=12$ intercepted between coordinate axes.

Solution:

Given line $3x+4y=12$

$\Rightarrow \frac{x}{4} + \frac{y}{4}=1$

line $l_2$ bisect $l_1$ at Q

$\therefore Q$ is the mid point

Q AB $\therefore$ Q $\left(\frac{4+0}{2},\frac{0+>}{2}\right)$

i.e. $Q (2,\frac{3}{2})$

Now line $l_2$ is $PQ$ passing through $P(3,1)$ & $Q (2,\frac{3}{2}).$

slope of $P Q=\frac{3 / 2-1}{2-3}=\frac{\frac{1}{2}}{-1}=-\frac{1}{2}$

Equation of PQ

$(y-1)=-\frac{1}{2}(x-3)$

$\Rightarrow 2 y-2=-x+3$

$\Rightarrow x+2 y-5=0 .$

By using the concept of equation of a line, that three points $(3,0), (-2,-2)$ and $(8,2)$ are collinear.

Solution:

Given three points $A(3,0), B(-2,-2)$ and $C(8,2)$

Equation of $AB, (y-0) = \frac{0+2}{3+2} (x-3)$

$\Rightarrow y=\frac{2}{5} (x-3)$

$\Rightarrow 5y = 2x-6 \Rightarrow 2x-5y-6=0$

put $x = 8$ & $y = 2$ in the equation of line

$2 \times 8 - 5 \times 2 - 6= 16-16=0$

$\therefore$ Hence C(8,2) satisfied the $Q \leqq$

$\therefore$ C(8,2) lies on the line.

Find the equation of the line passing through $(1,2)$ and making angle of $3^0$ with y-axis.

Solution:

Line $l$ makes $30^0$ with y-axis.

$\Rightarrow l$ makes $60^0$ with x-axis

$\therefore$ slope = tan $60^0 =\sqrt{3}$

$\therefore$ Equation fo $l$ passing through P(1,2) is

$(y-2) = \sqrt {3} (x-1)$ $\Rightarrow \sqrt{3}x-y+2-\sqrt{3}=0$

The perpendicular from the origin to the line $y= mx+c$ meets it at the point $(-1,2)$. Find the values of m and C

slope of OP $=\frac{0-2}{0+1}=-2\left(m_1\right)$

Since OP perpendicular l is $y=m x+c$

$\therefore$ slope of $l$ = $-\frac{1}{m_1} = \frac{1}{2} (m_2)$

$\therefore$ Equation of line $l$

$(y-2) = \frac{1}{2} (x+1)$

$\Rightarrow 2y-4=x+1$

$\Rightarrow 2y = x+5$

$\Rightarrow y=\frac{1}{2}x + \frac{5}{2}$

$\Rightarrow m=\frac{1}{2}$ & $c =\frac{5}{2}$

The points $P(1,2)$ and $R(0,-1)$ are two opposite vertices of a rhombus $PQRS$, find the equation of the diagonal $QS.$

Solution:

In the fig. O is the mid point of PR.

$\therefore$ O $\left(\frac{1+0}{2}, \frac{2-1}{2}\right)$ i.e. O $\left(\frac{1}{2},\frac{1}{2}\right)$

slop of PR (m) = $\frac{2+1}{1-0}=3$

slope of QS = $-\frac{1}{3}$

Equation of disapointed QS $\left(y-\frac{1}{2}\right)=-\frac{1}{3}\left(x-\frac{1}{2}\right)$

$\Rightarrow \frac{2y-1}{2} = -\frac{1}{2}\left(\frac{2y-1}{2}\right)$

$\Rightarrow -6y+3=2x-1 \Rightarrow 2x+6y-y=0$

$x+3y-2=0$